chapter-3时间序列.doc

CHAPTER 3MODELING VOLATILITY Answers to Questions1. Suppose that the et sequence is the ARCH(q) process et = vt . a0 +a1(et-1)2 +.+ aq (et-q)21/2Show that has the same form as the conditional variance of (3.1).Answer: Regardless of whether or not the actual regression residuals are used, the expected value of E(et)2 is:Et-1(et)2 = a0 + a1E(et-1)2 + . + aqE(et-q)2 Now using (3.8), et = vt . a0 +a1(et-1)2 +.+ aq (et-q)21/2 so that:Et-1(et)2 = Et-1(vt)2 (a0 + a1 (et-1)2 +.+ aq (et-q)2) = Et-1(vt)2 . Et-1a0 + a1 (et-1)2 +.+ aq (et-q)2 = a0 + a1 (et-1)2 +.+ aq (et-q)2Thus, using either (3.1) or (3.8):Et-1(et)2 = a0 + a1(et-1)2 +.+ aq (et-q)22. Consider the ARCH-M model represented by equations (3.23) through (3.25). Recall that et is a white noise disturbance; for simplicity let E(et)2 = E(et-1)2 = . = 1. A. Find the unconditional mean: Eyt. How does a change in d affect the mean? Using the example of section 6, show that changing b and d from (-4, 4) to (-1, 1) preserves the mean of the yt sequence. Answer: Combine (3.23), (3.24) and (3.25) and take the expectation of the result to obtain: Eyt = E (mt + et) = E (b + d.ht + et) = E b + da0 +ai (et-i)2 + etSince E(et)2 = E(et-1)2 = . = 1, and Eet = 0 it follows that: Eyt = b + d.(a0 +ai)Increasing b has a 1:1 effect on the mean; the effect of a one-unit change in d in Eyt is a0 + Sai. It is straightforward to show that a0 + Sai = 1. Since E(et)2 = E(vt)2ht = 1 and E(vt)2 = 1, it follows that E(ht) = 1. Hence: Eht = Ea0 + Sai (et-i)2 = a0 + Sai = 1. Given b = -4, d = 4, the unconditional mean of yt is:Eyt = -4 + 4 (a0 + Sai) = -4 + 4.1 = 0.Similarly, if b = -1 and d = 1, then Eyt = 0. Hence, changing b and d from (-4, 4) to (-1, 1) preserves the mean of the yt sequence.B. Show that the unconditional variance of yt when ht = a0 + a1 (et-1)2 does not depend on b, d, and a0.Answer: From Equation (3.23), yt = mt + et so that the unconditional variance of yt is: Var(yt) = var(mt) + var(et) + 2 Cov(mt,et). Note that: Var(et) = E(et)2 - E(et)2 = 1 - 02 = 1. Next, form cov(mt,et) as:Cov(mt,et) = E (mtet) - E(mt) . E(et) = E (b+d.ht) . et - E (b+d.ht) . 0 = E b+d . (a0 +a1(et-12) .et = E b.et+d . (a0 +a1(et-1)2) et = b.E(et)+d . E(a0 +a1(et-1)2 )et = b.0+d.0 = 0Now, find Var(mt) as:Var(mt) = var(b+d.ht) = d2var(ht) = d2vara0 +a1(et-1)2 = d2(a1)2var(et-1)2.Hence, the unconditional variance of yt does not depend on b and a0 since var(yt) = var(mt + et) and cov(ut, et) = 0. Thus:Var(yt) = 1 + d2(a1)2 . var(et-1)2However, as opposed to the assertion in the question, increasing the absolute value of d increases var(yt).3. Bollerslev proved that the ACF of the squared residuals resulting from theGARCH (p, q) process represented by (3.9) acts as an ARMA (m, p) process where m = max(p, q). You are to illustrate this result using the examples below.A. Consider the GARCH(1,2) process: ht = a0 + a1(et-1)2 + a2 (et-2)2 + b1 ht-1. Following the steps in the question, it is possible to write: (et)2 = a0 + (a1 + b1) (et-1)2 + a2 (et-2)2 - b1ht-1 + htwhere: ht = (et)2 - ht Show that ht is serially uncorrelated and that the (et)2 sequence acts as an ARMA(2,1) process.Answer: Recall that: et = vtwhere E(vt) = 0 and Var(vt) = 1. To show that ht is serially uncorrelated, we need to prove Cov(ht, hs) = 0, for ts. By definition, ht = (et)2 - ht, and et = so that:ht = (et)2 - ht = (vt)2ht - ht = (vt)2 - 1 ht.Since vt and ht are independent, it follows that:E(ht) = E(vt)2 - 1 ht= E (vt)2-1E(ht) = (1-1) . E(ht) = 0, t.As such, cov(ht,hs) = E ht-E(ht)hs-E(hs) = E (ht.hs) = E (vt)2-1ht.(vs)2-1hs cov(ht, hs) = E(vt)2-1E(ht).E(vs)2-1E(hs) = 0Thus, ht is serially uncorrelated. To show that the (et)2 sequence acts as an ARMA(2,1) process use the results that:E(ht) = 0 andCov (ht,hs) = 0, for ts.Since (et)2 = a0 + (a1+b1) (et-1)2 + a2 (et-2)2 - b1ht-1 + ht, it is immediately clear that the (et)2 sequence acts as an ARMA(2,1) process with autoregressive coefficients (a1+b1), a2, a moving average coefficient -b1, and ht as a white noise process.B. Consider the GARCH(2, 1) process: ht = a0 + a1(et-1)2 + b1 ht-1 + b2 ht-2. Show that it is possible to add ht to each side to obtain:(et)2 = a0 + a1(et-1)2 + b1 ht-1 + ht + b2 ht-2. Show that adding and subtracting the terms b1(et-1)2 and b2 (et-2)2 to the right-hand-side of this equation yields an ARMA(2,2) process.Answer: Given the GARCH(2, 1) process, ht = a0 + a1(et-1)2 + b1 ht-1 + b2 ht-2, it is possible to add ht = (et)2 - ht to each side to obtain:ht + (et)2 - ht = a0 + a1(et-1)2 + b1 ht-1 + b2ht-2 + (et)2 - ht. Hence: (et)2 = a0 + a1 (et-1)2 + ht + b1 ht-1 + b2 ht-2. Now, add and subtract the terms b1(et-1)2 and b2(et-2)2 to the right-hand-side of above equation yields:(et)2 = a0 + a1(et-1)2 + ht + b1ht-1 + b2ht-2 = a0 + a1(et-1)2 + ht + b1 ht-1 - b1(et-1)2 + b1(et-1)2 + b2ht-2 - b2(et-2)2 + b2(et-2)2 = a0 + a1(et-1)2 + ht - b1(et-1)2 - ht-1 + b1(et-1)2 - b2 (et-2)2 - ht-2+ b2(et-2)2 (et)2 = a0 + (a1 + b1)(et-1)2 + b2(et-2)2 + ht - b1ht-1 - b2ht-2Since ht acts as a white noise process, the (et)2 sequence acts as an ARMA(2,2) process with AR coefficients (a1 + b1) andb2 and MA coefficients (-b1), and (-b2).C. Provide an intuitive explanation of the statement:The Lagrange Multiplier for ARCH errors test cannot be used to test the null of white noise squared residuals against an alternative of a specific GARCH(p,q) process.Answer: Every GARCH(p,q) process has a representation ofARMA(m,p) process for the ACF of the squared residuals, where m=max(p,q). So that the Lagrange Multiplier for ARCH errors test cannotbe used to test the null of white noise squared residuals againstan alternative of a specific GARCH(p,q) process.D. Sketch the proof of the general statement that the ACF of the squared residuals resulting from the GARCH(p,q) process represented by (3.9) acts as an ARMA(m,p) process where m=max(p,q).Answer: Consider the GARCH(p,q) process: ht = a0 + ai(et-i)2 +bjht-j. Add the expression (e2t - ht) to each side to give:(et)2 = a0 +ai(et-i)2 +bjht-j + (et)2 - ht = a0 +ai(et-i)2 + bj(et-j)2 -bj(et-j)2 +bjht-j + (et)2 - ht = a0 + ai (et-i)2 +bj(et-j)2 -bj(et-j)2 - ht-j + (et)2 - ht = a0 + (a + b1)(et-1)2 + (a2 + b2) (et-2)2 + . + (am + bm)(et-m )2 -b1ht-1 - b2ht-2 -.-bp ht-p + ht,where ht (et)2 - ht, m = max(p,q), ai 0 for i q, and bj 0 for j p. Hence: (et)2 = a0 +(ak + bk)(et-k)2 + ht - b1ht-1 - b2ht-1 -b2ht-2 -.- bp ht-p, where m = max(p,q), ai 0 for i q, bj 0 for j p.Since ht is serially uncorrelated the ACF of the squared residuals resulting from the GARCH(p, q) process acts as an ARMA(m, p) process where m = max(p, q).4. Let y0 = 0 and let the first five realizations of the et sequence be: (1,-1,-2, 1, 1). Plot each of the following sequences:Model 1: yt = 0.5yt-1+etModel 2: yt = et - (et-1)2Model 3: yt = 0.5yt-1 + et - (et-1)2Answer: The time paths of the three models are shown in Figure 3M-1. Note that Model 3 contains autoregressive and ARCH-M effects. The effect of introducing the autoregressive term in going from Model 2 to Model 3 increases the volatility of the series. Similarly, the effect of introducing the ARCH-M effect in going from Model 1 to Model 3 is to increase the volatility of the series. The autoregressive coefficients and ARCH-M effects interact.B. For each of the three models, calculate the sample mean and variance of yt.Answer: For Model 1, the realizations are: (1,-0.5,-2.25,-0.125, 0.9375). The sample mean of Model 1 is: (1/5) (1 - 0.5 - 2.25 -0.125 + 0.9375) = -0.1875, and the sample variance is: (1/5) 12 +(-0.5)2+(-2.25)2 +(-0.125)2+(0.9375)2 - (-0.1875)2 = 1.40625.For Model 2, the realizations are: (1,-2,-3,-3, 0). The sample mean of Model 2 is: (1/5) (1-2-3-3+0) = -1.4, and the sample variance is (1/5) 12 +(-2)2+(-3)2+(-3)2+0 - (-1.4)2 = 2.64.For Model 3, the realizations are: (1,-1.5,-3.75,-4.875, -2.4375). The sample mean is (1/5) (1-1.5-3.75-4.875-2.4375) = -2.3125 and the sample variance is: (1/5) 12+ (-1.5)2 + (-3.75)2 + (-4.875)2 + (-2.4375)2 - (-2.3125)2 = 4.05625.Hence:Model Mean Variance 1 -0.1875 1.40625 2 -1.4 2.64 3 -2.3125 4.056255. The file labeled ARCH.XLS contains the 100 realizations of the simulated yt sequence used to create the lower right-hand panel of Figure 3.10. The following program will reproduce the reported results.Sample Program for RATS Users:all 100;* allocates space for 100 observationsopen data a:arch.xls;* opens the data set assumed to be on drive a:data(format=xls,org=obs) table / y;* produces the summary statistics for the ARCH(1) series ;* ARCH(1) series labeled y* Next, estimate an AR(1) model without an intercept and produce the ACF and PACF.boxjenk(ar=1) y / residscor(partial=pacf,qstats,number=24,span=4,dfc=1) resids* Now, define sqresid as the squared residuals from the AR(1) model and construct the ACF* and PACF of these squared residuals. set sqresid = resids*residscor(partial=pacf,qstats,number=24,span=4,dfc=1) sqresidlinreg sqresid;* estimate an AR(1) model of the squared residuals# constant sqresid1compute trsq= %nobs*%rsquared;* Calculate TR2 and its significance levelcdf chisqr trsq 1linreg sqresid;* estimate an AR(4) model of the squared residuals.# constant sqresid1 to 4nonlin b1 a0 a1 ;* prepares for a non-linear estimation b1 a0 and a1frml regresid = y - b1*y1 ;* defines the residualfrml archvar = a0 + a1*regresid(t1)*2;* defines the variancefrml archlogl = (v=archvar(t), -0.5*(log(v)+regresid(t)*2/v) ;* defines the likelihoodboxjenk(ar=1) y;* estimate an AR(1) in order to obtain an initialcompute b1=%beta(1) ;* guess for the value of b1 and a0compute a0=%seesq, a1=.3;* the initial guesses of a0 and a1* Given the initial guesses and the definition of archlogl, the next line performs the non-* linear estimation of b1, a0, and a1.maximize(method=bhhh,recursive,iterations=75) archlogl 3 *6. The file QUARTERLY.XLS contains the quarterly values of the U.S. Producer Price Index (PPI) that were used in Section 4. The following program will estimate the various GARCH models of the logarithmic change in the PPI. cal 1960 1 4;* Sets the calendar dates to begin in 1960:Q1all 2002:1;* through 2002:Q1open data a:quarterly.xls;* Change this line if the data is not on drive a: data(format=xls,org=obs)* The next three lines form the inflation rate as the logarithmic change in the PPIdif ppi / dylog ppi / lydif ly / dly*The next line estimate the ARMA(1, |1,4|) model and stores the residuals in the series resids. The * following lines creates the correlations and Q-statistics of resids.boxjenk(define=eq1,constant,ar=1,ma=|1 , 4 |) dly / residscor(qstats,number=24,span=4) resids* Next, create the squared residuals. The correlations of the squared residuals and the Q-statistics are * obtained using set r2 = resids*2cor(qstats,span=4,number=12) r2* Estimate a linear regression in the form of (3.17) using four lags of the squared residuals lin r2 ; # constant r2 1 to 4* Now use eight lagslin(nopri) r2 ; # constant r21 to 8* To exclude lags 5 through 8 use:exc ; # r25 to 8* The following block of instructions can be used to estimate the ARMA(1,(1,4) model with * ARCH(4) errors. Additional details can be found in Chapter 1 of the Programming Manual.set u = 0.0nonlin b0 b1 b2 b3 a0 a1 a2 a3 a4 frml e = dly - b0 - b1*dly1 - b2*u1 - b3*u4frml var = a0 + a1*e1*2 + a2*e2*2 + a3*e3*2 + a4*e4*2frml L = (u = e), -.5*(log(var)+e(t)*2/var)boxjenk(noprint,constant,ar=1,ma=|1,4|) dlycompute b0=%beta(1), b1=%beta(2), b2 = %beta(3) , b3 = %beta(4)compute a0=%seesq, a1=.1 , a2 = .1 , a3 = .6 , a4 = .1maximize(iterations=75) L 6 * To constrain the coefficients in the ARCH(4) process to be 0.4, 0.3, 0.2 and 0.1, useset u = 0.0nonlin b0 b1 b2 b3 a0 a1frml e = dly - b0 - b1*dly1 - b2*u1 - b3*u4frml var = a0 + a1*(.4*e1*2 + .3*e2*2 + .2*e3*2 + .1*e4*2)frml L = (u = e), -.5*(log(var)+e(t)*2/var)boxjenk(noprint,constant,ar=1,ma=|1,4|) dlycompute b0=%beta(1), b1=%beta(2), b2 = %beta(3) , b3 = %beta(4)compute a0=%seesq, a1=.1 , a2 = .1 , a3 = .6 , a4 = .1maximize(iterations=75) L 6 * To estimate the ARMA(1,(1,4) with GARCH(1,1) errors, useset w = 0.0set u = 0.0nonlin b0 b1 b2 b3 a0 a1 a2frml e = dly - b0 - b1*dly1 - b2*u1 - b3*u4frml var = a0 + a1*e1*2 + a2*w1frml L = (u = e), (w = var), -.5*(log(var)+e(t)*2/var)boxjenk(noprint,constant,ar=1,ma=|1,4|) dlycompute b0=%beta(1), b1=%beta(2), b2 = %beta(3) , b3 = %beta(4)compute a0=%seesq, a1=.3 , a2 = .5maximize(iterations=75) L 6 *The residuals are contained in the series u. You can perform the appropriate diagnostics on this series. 7. The second series on the file ARCH.XLS contains 100 observations of a simulated ARCH-M process. The following programs will produce the indicated results.* The next three lines read in the 100 observations from the file. all 100open data a:/arch.xlsdata(format=xls,org=obs)table / y_m ;* The second series on the file is called y_m. TABLE produces the desired ;* summary statistics. * The following instruction produces the graph of the ARCH-M processgraph(header=Simulated ARCH-M Process) 1 ; # y_m* To estimate the MA(|3,6|) process and save the residuals as resids, use:boxjenk(constant,ma=|3,6|) y_m / resids*The correlations of resids (and the Q-statistics) are given bycor(partial=pacf,qstats,number=24,span=8) resids* Now form the squared residuals and obtain the autocorrelations usingset ressq = resids*resids ;* Form the squared residualscor(partial=pacf,qstats,number=24,span=4) ressq * Perform the Lagrange multiplier test for ARCH(4) errors by regressing the squared error on its own* four lags.linreg ressq; # constant ressq1 to 4*To obtain the F-statistic useexclude ; # ressq1 to 4* Alternatively, to obtain TR2 usecompute trsq = %nobs*%rsquared ;* Compute T*R2 and obtain the cumulative densitycdf chisqr trsq 4 ;* of trsq as chi-square with 4 degrees of freedom.set u = 0.0nonlin b0 b1 a0 a1frml var = a0 + a1*u1*2;* variance equationfrml e = y_m - b0 - b1*var(t);* mean equationfrml L = (u = e), -.5*(log(var)+e*2/var);* likelihood functioncompute b0=.8, b1=.6 , a0=.2, a1=.7;* initial guessesmaximize L 2 *8. Consider the ARCH(2) process: Et-1(et)2 = a0 + a1(et-1)2 + a2(et-2)2.A. Suppose that the residuals come from the model yt = a0+a1yt-1+et. Find the conditional and unconditional variance of yt in terms of the parameters a1, a0, a1, and a2.Answer: To find the conditional variance of yt, first note that the conditional mean of yt is: Et-1yt = Et-1(a0+a1yt-1+et) = a0+a1.yt-1. The conditional variance of yt is:Var(yt|yt-1, yt-2,.) = Et-1yt-Et-1(yt)2 = Et-1 (a0+a1yt-1+et) - (a0+a1yt)2 = Et-1(et)2 = Et-1a0 + a1(et-1)2 + a2(et-2)2 = a0 + a1(et-1)2 + a2(et-2)2To find the unconditional variance of yt find the particular solution for yt as: yt = a0/(1-a1) + et + a1et-1 + (a1)2et-2 + . Thus, the unconditional mean is: Eyt = a0/(1-a1). Using the solution for yt and Eyt, the unconditional variance of yt is: Var(yt) = E yt-E(yt)2 = Eet + a1et-1 + (a1)2et-2 + (a1)3et-3 + . 2= s2/1 - (a1)2Given that et = vt., it follows that (et)2 = v2t a0+a1(et-1)2 + a2(et-2)2). Since Eet = Evt = 0:E(et)2 = E (vt)2Ea0 + a1(et-1)2+a2(et-2)2 = a0+a1 E(et-1)2 + a2 E(et-2)2Since the unconditional variance of et is identical to that of et-1 and et-2, the unconditional variance is:E(et)2 = a0 /(1-a1-a2) so that the unconditional variance of yt is:Var(yt) =(a1)2iVar(et-i) = a0 /(1-a1-a2)(1-(a1)2).B. Suppose that yt is an ARCH-M process such that the level of yt is positively related to its own conditional variance. For simplicity,let: yt = a0 + a1(et-1)2 + a2(et-2)2 + et. Trace out the impulse response function of yt to an et shock. You may assume that the system has been in long-run equilibrium (et-2 = et-1 = 0) but now e1 = 1. Thus, the issue is to find the values of y1, y2, y3, and y4 given that e2 = e3 =.= 0.Answer: Iterating forward from the initial conditions: y1 = a0 + a1(e0)2 + a2(e-1)2 + e1 = a0 + a1.02 + a2.02 + 1 = a0+1y2 = a0 + a1(e1)2 + a2(e0)2 + e2 = a0 + a1 + a2.02 + 0 = a0+a1y3 = a0 + a1(e2)2 + a2(e1)2 + e3 = a0 + a1.02 + a2.12 + 0 = a0+a2y4 = a0 + a1(e3)2 + a2(e2)2 + e4 = a0 + a1.02 + a2.02 + 0 = a0y5 = a0 + a1(e4)2 + a2(e3)2 + e5 = a0 + a1.02 + a2.02 + 0 = a0,.Hence:dy1/de1 = 1; dy2/de1 = a1; dy3/de1 = a2; and dyi/de1 = 0 for i 3.C. Use your answer to Part B to explain the following result. A student estimated yt as an MA(2) process and found the residuals to be serially uncorrelated. A second student estimated the same series as the ARCH-M process: yt = a0 + a1(et-1)2 + a2(et-2)2 + et. Why might both estimates appear reasonable? How would you decide which is the better model?Answer: If a1 and a2